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Let $k$ be a field and let $k[t, x]$ be the polynomial ring in two variables. Is it true that $k[t][[x]]=k[t]\otimes_kk[[x]]$ or does the completion of $k[[x]]$ make that impossible? We can compute

$$k[t][[x]]\cong\lim_i k[t,x]/(x^i)=\lim_i\oplus_{j\leq i}k[t]x^j=\prod_i k[t]x^i=\\ \prod_i k[t]\otimes_k x^i,$$

where $x^i$ in the tensor product is the $1$-dimensional vector space over $k$ spanned by $x^i$.

The computation does not tell me much about whether we can commute $k[t]$ with the direct product. Can we?

Flavius Aetius
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    Consider $\sum_{n=0}^\infty t^nx^n$. Does it belong to $k[t][![x]!]$, or $k[t]\otimes_kk[![x]!]$? – Yai0Phah Oct 16 '22 at 12:21
  • Also what do you think of $k[[x]][t]$? – reuns Oct 16 '22 at 12:35
  • @Yai0Phah great example! – Kenta S Oct 16 '22 at 13:17
  • @Yai0Phah, Ok, you got me convinced with this example. $\sum_{n\geq0}t^nx^n$ belongs to $k[t][[x]]$, but not to $k[t]\otimes_kk[[x]]$. However, there is still an injective embedding $k[t]\otimes_kk[[x]]\to k[t][[x]]$ coming from the completion, correct? In particular, we have $k[t]\hat{\otimes}_k k[[x]]\cong k[t][[x]]$, correct? – Flavius Aetius Oct 16 '22 at 13:30
  • @reuns, I do not understand t your question. Are you trying to say that $k[t][[x]]\neq k[[x]][t]$? If yes, I would disagree with this. – Flavius Aetius Oct 16 '22 at 13:33
  • $k[t][[x]]$ and $k[[x]][t]$ are indeed different rings. Let $A=k[t],B=k[[x]]$ then one is $B[t]$, polynomials in $t$, the other is $A[[x]]$, containing $\sum_{n\ge 0} t^n x^n$. – reuns Oct 16 '22 at 14:07
  • So, we have $k[[t]]\hat{\otimes}_kk[[x]]\cong k[[t,x]]\cong k[[t]][[x]]\cong k[[x]][[t]]$, but the statement is not true when one of the modules is not complete, that is, $k[t]\hat{\otimes}_k k[[x]]\neq k[t][[x]]$? Is this the statement? – Flavius Aetius Oct 16 '22 at 14:25
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    I explained to you several times that $\hat{\otimes}$ is ambiguous. So what is the meaning of $k[t]\hat{\otimes}_k k[[x]]$ to you. – reuns Oct 16 '22 at 14:31
  • $k[t]\hat{\otimes}_k k[[x]]$ denotes the completion of the algebraic tensor product with respect to the ideal $(x)$ in $k[[x]]$. There are no open complex vector subspaces of $k[t]$, so $k[t]\hat{\otimes}_kk[[x]]$ is to be understood the same way as a $V\hat{\otimes}_kk[[x]]$ for some complex vector space $V$. – Flavius Aetius Oct 16 '22 at 15:17
  • I always use this definition of a completion of a tensor product: https://stacks.math.columbia.edu/tag/0AMU – Flavius Aetius Oct 16 '22 at 15:19
  • Why don't you mention systematically to which topoogy (metric or adic) you are completing? $k[t]\otimes k[[x]] \cong k[[x]][t]$ and so $\varprojlim k[t]\otimes k[[x]]/(x^n)\cong\varprojlim k[[x]][t]/(x^n)\cong \varprojlim k[x][t]/(x^n)\cong \varprojlim k[t][x]/(x^n)\cong k[t][[x]]$ – reuns Oct 16 '22 at 18:16
  • Right, I complete with respect to the $(x)$-adic topology. I did not consider the so-called metric topology because I do not know how it is defined. You show that $k[[x]][t]\cong k[t]\otimes_k k[[x]]$ and $k[t][[x]]\cong\lim_n k[t]\otimes_kk[x]/(x^n)$ from which it is evident that $k[t][[x]]\neq k[[x]][t]$. – Flavius Aetius Oct 16 '22 at 22:47
  • @reuns, I posed a question here:https://math.stackexchange.com/questions/4555264/completion-of-a-scalar-extension-in-the-metric-topology that you might find useful (or not). If you answer it, I intend to accept it. – Flavius Aetius Oct 18 '22 at 10:55

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As described in the comments, $k[t] \otimes k[[x]]$ is a proper subspace of $k[t][[x]]$, and the issue is that the tensor product only contains elements which have bounded degree in $t$, so cannot contain e.g. $\sum t^n x^n$.

Abstractly the issue is that, because $k[t]$ is infinite-dimensional, taking the tensor product $k[t] \otimes (-)$ does not commute with infinite products. We can show more generally that if $M$ is a flat module then $M \otimes (-)$ commutes with infinite products iff $M$ is finitely presented; see this math.SE answer.

Probably taking the completion fixes this but I'm not familiar enough with completions to say this with any real confidence.

Qiaochu Yuan
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  • Thank you, Qiaochu Yuan, for your answer! – Flavius Aetius Oct 16 '22 at 22:49
  • Qiaochu Yuan. Maybe the assumption that $M$ is flat is not necessary at all. Have a look at the answer by username “marlu”: https://math.stackexchange.com/questions/1916457/direct-product-commutes-with-tensor-product. The flatness assumption is completely omitted and it does not play a role in the proof. – Flavius Aetius Nov 12 '22 at 01:15
  • According to your answer in the referred link, $M\otimes_A(-)$ commutes with limits if and only if it commutes with infinite products. That implies that $M\otimes_A(-)$ commutes with limits if and only if it is flat and finitely presented. But in light of “marlu’s” answer, finite presentation of $M$ suffices, correct? – Flavius Aetius Nov 12 '22 at 02:08
  • Aha. You are saying that when $M$ is flat, $M\otimes_A(-)$ commutes with limits if and only if $M\otimes_A(-)$ commutes with infinite products which is if and only if $M$ Is finitely presented. – Flavius Aetius Nov 12 '22 at 02:54
  • @Flavius: yes, that's what I meant. I wasn't able to figure out what happens without flatness. – Qiaochu Yuan Nov 12 '22 at 03:36
  • But you agree that for any $A$-module $M$, the functor $M\otimes_{A}(-)$ commutes with infinite products if and only if $M$ Is finitely presented, right? – Flavius Aetius Nov 12 '22 at 11:42
  • That seems plausible but I'm not familiar enough with the five lemma to easily check marlu's argument for myself. – Qiaochu Yuan Nov 15 '22 at 19:34
  • Here is a paper from 1972 in which $M$ does not need to be flat for $M\otimes_A(-)$ to commute with infinite products: https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-43/issue-1/Distributing-tensor-product-over-direct-product/pjm/1102959646.full. The statement is in Theorem L, b) on the $1$st page. – Flavius Aetius Nov 16 '22 at 11:57
  • I wonder If you do not need $M$ to be flat for $M\otimes_A(-)$ to commute with infinite products, then probably you do not need flatness for $M\otimes_A(-)$ to commute with inverse limits? Both are basically the same. – Flavius Aetius Nov 16 '22 at 11:58
  • I posted in https://math.stackexchange.com/questions/181004/inverse-limit-of-modules-and-tensor-product/4577954#4577954 a version of marlu's proof of the claim that $M\otimes_A(-)$ commutes with inverse limits if $M$ is finitely presented. It seems to me that there is no need to assume that $M$ is flat. – Flavius Aetius Nov 16 '22 at 14:26